Solution We’ll use the first method in this example to find the length. Recall that the length of the chord is given by |x1 – x2|\( \sqrt{1+m^2} \) .

By substituting the value of y from the equation of the line in that of the circle, and after simplification, we get 5x2 – 18x – 35 = 0.

The difference of the roots of the equation is given by |x1 – x2|= \( \sqrt{(18/5)^2 -4.7} \) = 32/5

And by plugging in the values in the formula, we get the required length as 8. As simple as that!

Example 2 The line 3x – 4y + 5 = 0 cuts the circle x2 + y2 = 10 at A and B. Find the length of the chord AB.

Solution This problem as exactly the same as above, except that we’ll be using the second formula that we derived in the previous lesson, i.e. 2\(\sqrt{r^2-d^2}\)

The distance of the line from the centre of the circle is given by |3(0) – 4(0) + 5|/5 which equals 1.

Now all we have to do is plug in the values. The required length equals 2\(\sqrt{10-1}\) = 6.

Example 3 Find the equation of the line, passing through (2, 3) which cuts off a chord of length 8 units on the circle x2 + y2 = 25.

Solution In this case we’ve been given the length of the chord, and are required to find its equation.

Let’s assume the slope of the chord to be ‘m’. Then the equation of the chord in the point-slope form will be y – 3 = m(x – 2). The distance of this chord from the center of the circle will be |3 – 2m|/\(\sqrt{1+m^2}\)

We know that the length of the chord is given by l = 2\(\sqrt{r^2-d^2}\) , which equals 8. Therefore the distance of the chord from the center or ‘d’ must be \(\sqrt{r^2-{l^2}/4}\) , which in this case will be 3.

On equating the value of ‘d’ with the expression obtained in m, we get two values of m: 0 and – 12/5

Therefore the required equations are: y = 3 and 12x + 5y = 39

And that’ll be all for this lesson. Hope you have a better understanding of how to obtain chord lengths using the equations of lines and circles.